Correction: the theorem is
h . either (f, g) = either (h . f, h . g)
(Thanks to Lennart for pointing out the typo.)
From: rj248...@hotmail.com
To: haskell-cafe@haskell.org
Subject: Clean proof?
Date: Sun, 23 May 2010 15:41:20 +
Given the following definition of either, from the
Lennart wasn't pointing out a typo. He was pointing out a fundamental
issue with such identities in a partial call-by-name language. If
h = const 42
then
(h . either (f, g)) undefined
evaluates to 42. But the evaluation of
(either (h . f, h . g)) undefined
is non-terminating.
On Sunday 23 May 2010 18:24:50, R J wrote:
Correction: the theorem is
h . either (f, g) = either (h . f, h . g)
Still not entirely true,
const True . either (undefined, undefined) $ undefined = True
while
either (const True . undefined, const True . undefined) undefined =
undefined
Actually, I didn't notice the typo. It's still not a true statement.
(h . either (f, g)) undefined /= (either (h . f, h . g)) undefined
Also, it's not exactly the function either from the Prelude.
-- Lennart
2010/5/23 R J rj248...@hotmail.com:
Correction: the theorem is
h . either
On Sun, May 23, 2010 at 11:38 AM, Daniel Fischer
daniel.is.fisc...@web.de wrote:
On Sunday 23 May 2010 18:24:50, R J wrote:
Correction: the theorem is
h . either (f, g) = either (h . f, h . g)
Still not entirely true,
const True . either (undefined, undefined) $ undefined = True
